Extreme operator-valued continuous maps
R. Grz~glewicz
Abstract. Let .oq~(H) denote the space of operators on a Hilbert space H. We show that the extreme points of the unit ball of the space of continuous functions
C(K, .LP(H))
(K-compact Hausdorff) are precisely the functions with extremal values. We show also that these extreme points are (a) strongly exposed if and only if dim H < ~ and card K<,~, (b) exposed if and only if / / is separable and K carries a strictly positive measure.1. Introduction
Let
C(K, X)
denote the Banach space o f all continuous functions from a c o m p a c t Hausdorff space K to a Banach space X equipped with the s u p r e m u m n o r m Ilfll=supk~x llf(k)]l. ByB(X)
we denote the unit ball of the Banach space X.There is a natural conjecture a b o u t extreme points:
( . )
fEextB(C(K,X))
if and only iff(k)EextB(X)
for allkEK.
Obviously the " i f p a r t " in ( . ) always holds. This conjecture has been proved to be true under various additional assumptions. We list some o f the known results.
T h e conjecture is true if:
1) X is strictly convex, 2)
B(X)
is polytope,3)
X=(C(K1))*=M(K1),
where K1 is a c o m p a c t H a u s d o r f f space ([31], T h e o r e m 4, 5). N o t e that in this case the conjecture ( . ) is equivalent to the fact that extreme operators in the unit ball o f the space o f c o m p a c t operatorsLa(C(K1), C(K))
are nice in the sense o f Morris and Phelps. F o r other similar results on the space o f weak* continuous functions from K into
(C(K1))*
(this corresponds to the case o f extreme point of the unit ball o f the space o f all bounded linear operatorss C(K))
(see [6], p. 490)) under various assumptions on K and Kx (see [2], [1], [34], [9], [20], see also [35], [36] for negative examples).4) X has 3.2.I.P. ([37]).
74 R. Grz~lewiez
5)
B(X)
is stable (to get it we use (iii')) in [4] and Michael's selection Theorem [29]).6) X=LX(#) ([39]).
7) X=L~'(It) is an Orlicz space equipped with the Luxemburg norm ([18]).
8)
X=M(Ka,Z)
is the space of Z-valued regular Borel measures of finite varia- tion, Z is a Banach space,/(1 is a compact Hausdorff space ([40]).The conjecture ( . ) is not true for every Banach space X. A negative example was given by Blumental, Lindenstrauss and Phelps for
X=R 4
equipped with a suitable norm ([2]). The class of finite dimensional spaces for which conjecture ( . ) holds was described by Papadopoulou ([32]). She proved thatB(X)
(dim X<oo) is stable if and only if all k-skeletons (k = 0, 1 ... n) of B (X) are closed (a k-skeleton ofB(X)
is a set of allxEB(X)
such that the face generated by x has dimension less than or equal to k).After dealing with the conjecture for Banach spaces the next natural step is to consider X as a space of linear operators. We denote by .LP(Y) the space of all bounded linear operators from a Banach space Y into itself equipped with the operator norm.
For X=Lz~ l < p < ~ o , p ~ 2 , the conjecture does not holds. Indeed, we have the following example. Let (a,)El v be such that II(an)Jlp=l and a , > 0 for all n. We define fEC([0, 1], .~~ by
f(k)=Tk,
kE[0, 1] whereTk((x.)) = x , ( ~ , ~/af + ag, ~c-~. [/o7 + ag, a a, a, .... ),
(x,)El p. We have for p > 2 ,
TkEextB(Le(IV))
for kE(0, 1) and TkqextB(~(/P)) for k = 0 or 1 (see [15], see also [14]). Using adjoint operators an analogous ex- ample can be written for pE(1, 2).In this note we consider the case p = 2 . It turns out that the conjecture ( . ) holds for the space of operators acting on a Hilbert space (Section 2).
In Section 3 we consider under what conditions elements of ext
B(C(K,
5e(H))) are exposed or strongly exposed. We show that extreme points ofB(C(K,
.oq'(H))) are (a) strongly exposed if and only if dim H < ~ and card K < ~o,(b) exposed if and only if H is separable and K carries a strictly positive measure.
2. Extreme points in
B(C(K,
(H)))Let H be a (real or complex) Hilbert space. We denote by PrE.oq'(H) the orthogonal projection onto a subspace E of H. The set of extreme points ext B (.oq' (H)) coincides with the set of all isometries and coisometries (see [23] for the complex case, and [16] for the real case). Our aim is to characterize ext
B(K, ~(H)).
Note that the below presented Theorem was proved in [14, p. 314, ( . . ) ] in the case when H is finite dimensional.Extreme operator-valued continuous maps 75 In the finite dimensional case the dual of LZ(H) was also considered, and it turns out that the unit ball of s is stable (see [17]), so the conjecture ( . ) also holds for C(K, B(LZ(H)*)) (dim H < ~).
Now we recall some facts we will use in the proof of Theorem 1 below.
Let TC s (H) be a contraction. We put
M ( T ) = {hEn: [IZhll = Ilhll},
We have Z*Th=h for h~M(T). Moreover T ( M ( T ) ) = M ( T * ) and T ( M • M•
Obviously, if T~extB(.,q'(H)) then neither T T * = I nor T * T = I i.e.
M • (T) # {0} # M • (T*).
Put S = ( ( I + T*T)/2) x/2. We have I/2<=S<=I. Therefore S -1, S 1/2 and S -x/~
exist. We have
0~$1/2<-L
Hence0 <= S~/2(2Sa/Z-I) = 2 S - S ~/2 <- L Since T*T<=S 2 we have
llTxII ~ = (T*Zx, x) <= (S2x, x) = !15x112.
So [ITS-1I[<_-I. We get
IITS-X/ell ~_ [ITS-lit [lSa/2ll <_- 1 and
[IT(2I-S-a/2)[I <-[IZS-l[I 1[2S-St/~l[ ~ 1.
Theorem 1. Let H be a Hilbert space and let K be a compact Hausdorff space. Then f ~ ext B(C(K, s (H)))
i f and only i f
f(k)~extB(LZ(H)) for all kCK.
Proof. Let f~B(C(K, .La(H))). Assume that f(ko) fails to be extremal for some koCK, i.e.
M • (f(ko)) # {0} # M • (f(ko)).
We need to prove that then
f r ext B(C(K, ~ (H))).
Put T k=f(k). We consider two cases.
_ - 1 l ~
1 ~ There exist ko~K and xCMX(Tko) such that Tk0X~0. Put A(k)--TkSk , and f2(k)=2Tk-TkS[XS~/2, where Sk=((I+Tk*Tk)/2) u2. Obviously
ae(H))) and f = (A+A)/2.
Since
M(Tko ) = M(T~oTko ) = M(S~o ) = M(S~o) = M(S~k/o ~)
76 R . G r z ~ l e w i e z
and
Tk, T~xEM (S~,)
* .t 11~ we haveSV2T,* T,
ko ko ko-~, <IIZ~Tkoxll.
Since
S~ 12
and Tk*T k commute, we get9 /2 *
TioTkoS~ko x ~ T~oTkox,
soTkoS~k~x ~ Tko x.
Hence
Tko~TkoS~l/*
andfl(ko)~f2(ko)
i.e. f is not extreme.2 ~
Tk(M•
= {0} =T~(M•
for allk~K.
Then Tk*Tk=P~(T~ ). Therefore
k-~PM(r, )
is a continuous function, sok-~PM•
is a continuous function, too. Fix
koEK
such thatf(ko)r
ChooseeEM • (f(ko))
andgEM •
(f(k0)*) with [[ei[ =[]gl] = 1. We haveIITk + PM•174 <= 1.
Hence f is not extreme.
3. Exposed and strongly exposed points in
B(C(K, (H)))
A point q0 in a convex set Q of a (real or complex) Banach space E is said to be exposed if there exists a bounded R-linear functional ~: E-~R such that ~ (q0) >
~(q) for all
qEQ\{qo}.
An exposed pointqoEQ
is called strongly exposed if for any sequence q.E Q the condition~(q.)~(qo)
implies llq.-q0]l-~0. Obviously each exposed point is extreme.Note that extreme points of
B(.~(H))
are strongly exposed in and only if H is a finite dimensional Hilbert space, and exposed but not strongly exposed if and only if H is separable infinite dimensional. Moreover there are no exposed points inB(.~(H))
if and only if H is not separable ([19]).Consider the Bochner LP-space ( l < p < oo). For
f~LP(ll, X) (X
is a Banach space), if I l f [ ] = l andf(t)/llf(t)llEextB(X)
for t~suppf/~-a.e., thenf C e x t
B(LP(I~, X)).
Generally the converse does not holds. A negative example was given by Greim [12]
for a nonseparable Banach space X. But for all separable Banach spaces X this property characterizes extreme functions (see [38], [21], [22], [13]). For the strongly exposed points the analogous natural condition on the values o f f are sufficient, whenever X is smooth ([10], see also [11]). Recently W. Kurtz considered strongly exposed points in Bochner--Orlicz spaces [25] (see also [26]). A compact Hausdorff space K is said to carry a strictly positive measure, if there exists a strictly positive Radon measure # on X (i.e. p ( q / ) > 0 for all non-empty open subsets q/ of X).
Extreme operator-valued continuous maps 77 Several authors have workes on the problem of the characterization of spaces X which carry a strictly positive measure. Maharam [28] has given necessary and sufficient conditions for the existence of strictly positive measures. Kelley considers strictly positive measures on Boolean algebras. In his work [24] he introduced the notion o f the intersection number of a collection of subsets to give a characteriza- tion of spaces which carry a strictly positive measure.
Next it turns out that the countable chain condition is not sufficient for the existence of such a measure (see Gaifman [8]). Note that in the case of a compact Hausdorff space, the problem mentioned above is equivalent to the problem of existence of a finitely additive strictly positive measure. Rosenthal ([33], Th. 4.5b) proved that C (X) carries a strictly positive functional if and only if its dual C (X)*
contains a weakly compact total subset. Other results can be found in [7], [30], [3].
We refer the reader to [5, Chapter 6] for a survey of known results about strictly positive measures. In fact, we can consider a strictly positive measure on X as a functional on
C(X)
which exposes the function 1.Theorem 2.
I f a Hilbert space H is separable and a Hausdorff compact space K carries a strictly positive measure, then each extreme point of B(C(K,
L~'(H)))is exposed.
I f H is not separable or K does not carry a strictly positi~'e measure then B(C(K, ~
(H)))contains no exposed points.
I f H is infinite dimensional then B(C(K,
L~~contains no strongly exposed points.
Proof.
Suppose that H is separable and K carries a strictly positive measure.Assume that /z(K)= 1. We fix an orthonormal basis
{ei}iet
in H. Fix foEextB(C(K,
A~Put K I = { k E K :
f(k)
is an isometry} andK 2 = K \ K I .
The set Kx is closed. Let(ai)tel
be a sequence o f strictly positive reals such that ~ i e t as= 1. We define a functional ~ onC(K, .~(H))
by( f ) = f r~ Z , c I a,
Re (If(k)] (e,), [fo (k)] (e,))
d/z+ f K% Z , e , a,
R e ([f(k)]*(e,),[fo(k)l*(e,)) d#,
fEC(K, .s
The functional exposesfo inB(C(K,
.oq~ Indeed ~(f)_-__l=~(fo) for all
fEB(C(K,
Le(n))).Suppose that ~ ( f x ) = 1 for some
fxEB(C(K,
La(H))). Because Re ([ fx (k)] (e,), [ fo (k)](el)) <= I,
the condition ~ ( f x ) = l implies that
[f~(k)](e,)=[fo(k)](e,)
for alliEI
andkEK~
78 R. Gr-z~lewiez
/z-a.e. Analogously f x = f o p-a.e, on Kz. Hence by continuity
fx=fo.
So we finish the proof of the first part of theorem.Now suppose that a functional ~0 exposes
foEextB(C(K,L#(H))) in B(C(K,
.La(H))). We define a functional m onC(K)
byre(h) = ~o(hfo), hEC(K).
We claim that m is strictly positive. Indeed, suppose to get a contradiction, that there exists
hoEC(K )
such that 0 ~ h o ( k ) ~ l , ho~0, and m(h0)<0.Then
~o((1-ho)fo) = m(1--ho) -<-- re(l) = r
and hofo ~ 0 , which is impossible. It follows that K carries a strictly positive measure.
Consider now a function n on all subsets of 1 defined by
n(L) = ~o(foPl~ te,: ~ L))
L C I
(in the case when allfo(k)
are coisornetries we define n by n ( L ) =~0(P~(~,:~cz~f0)). The function n is finitely additive on the family of all subsets o f / . Moreover n ( I ) = l and
n(L)~O
forL c I .
Suppose that n(L0)--0 for some non-empty L 0 c I . ThenSO
~0(f0~Tte,:~L0 }) = 0,
(i.e. 40 does not expose fo). This contradiction proves that n(L0)>0. Hence I is countable and H is separable, which yields the second part of the theorem.
Let (Li)j~ N be a sequence of non-empty disjoint subsets o f L Then
n ( L j ) ~ O.
Hence
~0(f0e~e ~CLg) = ~o(fo)--n(Lj) 1, ~offo)
and
Ilf0-foe~t,,:~r =
IlfoP~ te,:ie Lsll I = 1.
Therefore fo is not strongly exposed in the case when I is infinite (i.e. H is infinite dimensional). This proves the third part of the theorem.
Theorem 3.
Let H be a finite dimensional Hilbert space and K be a Hausdorff compact space. Then each extreme point of B(C(K,
.L#(H)))is strongly exposed, if and only if K is finite.
Proof.
Let d i m H < ~ o and c a r d K < r FixfoEB(C(K,
.La(H))). Let r/k be a functional on .L#(H) which strongly exposesfo(k),kEK.
It is easy to see that aExtreme operator-valued continuous maps 79
functional ~ defined by
r = Z k r f ~ C ( K , .C#(H)), exposes ./~.
N o w suppose that card K = ~o. Let a fun ction al 3o expose f0 C ext B (C (K, .C>e (H))).
Let (k.).(N be a sequence o f distinct points o f K such that lira k . = k o . Let h . E C ( K ) , nEN, be such that 0-<__h. <- I, h . ( k . ) = 1 and supp h. c~supp h . = 0 if n1#n2. Put a#=r N o t e that a.=>0 because
~ 0 ( f 0 ) - a. = 3 0 ( ( 2 - h . ) f 0 ) ~ ~0(f0).
F o r every finite subset L o f N we have
Z . e z an = r h.)fo) ~- r
Hence a.---O. Therefore
~ o ( ( 1 - h , ) f 0 ) = ~o(fo)-a~ -~ ~o(fo) and
1I(2 -h.)f0-f011 = l i h . f 11 = 2.
Thus fo is not strongly exposed by 40. This ends the proof.
Remark. I would like to express my thanks to the referee for his useful remarks which shorten the p r o o f o f T h e o r e m 2.
Acknowledgement. Written partially while the author was a research fellow of the Alexander von Humboldt-Stiftung at the Mathematisches Institut der Eberhard- Karls-Universitiit in Tiibingen.
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Received September 26, 1986
Received in revised form November 9. 1989
R. Gr7~tlewicz
Institute of Mathematics Technical University Wyb. Wyspiafiskiego 27
50-370 Wroclaw Poland